1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Reversed limit definition for monotonic functions

  1. Jul 28, 2017 #1
    1. The problem statement, all variables and given/known data

    Does the delta-epsilon limit definition in reverse work for describing limits in monotonic functions?

    By reversed, one means for

    lim (x -> a) f(x) = L

    if for each δ there corresponds ε such that

    0 < | x-a | < δ whenever | f(x) - L | < ε.

    2. Relevant equations


    3. The attempt at a solution

    I am thinking that it works, because this definition means that the range interval must lie within the domain interval, and it can be seen that shrinking δ also shrinks ε, which is how the usual definition works but in reverse.

    I don't think this would work for non-monotonic functions because there can be many f(x) that satisfy
    | f(x) - L | < ε but not | f(x) - L | < ε. Hopefully someone can also confirm this part too.

    Thanks
     
    Last edited: Jul 28, 2017
  2. jcsd
  3. Jul 28, 2017 #2

    fresh_42

    Staff: Mentor

    You can define whatever you like, as long as it is well defined. For this reason it should be helpful to indicate which variable depends on which and to explicitly write the quantors. If I read it correctly, then ##\varepsilon = \varepsilon(\delta)##. But then I have difficulties with the next line. I read ##\forall \, \delta \, \exists \, \varepsilon(\delta)\, : \,\vert \, f(x)-L\,\vert \, < \varepsilon(\delta) \Longrightarrow \,\vert \,x-a\,\vert \,< \delta \,##. The order of the quantors don't seem to reflect to the order of the conclusion. And how do you specify ##L##\,?

    The remaining question is in any case: What for?
     
  4. Jul 28, 2017 #3
    I forgot the part where δ > 0 and ε > 0, so I think it would be written like this:

    ##\forall \, \delta \ > 0, \exists \, \varepsilon(\delta)\ > 0, : \,\vert \, f(x)-L\,\vert \, < \varepsilon(\delta) \Longrightarrow \, 0 < \vert \,x-a\,\vert \,< \delta \,##

    L is the supposed limit.

    lim (x -> a) f(x) = L

    I'm not very experienced in this type of math, so it's just a random inquiry.
     
  5. Aug 1, 2017 #4

    fresh_42

    Staff: Mentor

    It doesn't really make sense. You can have ##f(x) = L## for several points ##x## which can be far away from ##x=a##. That's where "what for" is needed. The way it is written now, is only confusing (IMO). It is more a condition of monotone behavior than on limits, because it says, ##\lim_{x \to a}f(x)=L## can only happen in a neighborhood of ##x=a## and nowhere else.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Reversed limit definition for monotonic functions
  1. Monotonic function (Replies: 2)

  2. Monotonic function (Replies: 8)

Loading...